Age Relative? Time, Speed and Paradox Explored

In summary, the conversation discusses the concept of time dilation in relation to the Special Theory of Relativity. It explains how time is relative and can be affected by speed, causing a person on a spaceship traveling close to the speed of light to age slower than someone on Earth. The conversation also delves into the twin paradox, where one twin ages slower after a journey in space, and the resolution of this paradox using the relativity of simultaneity. It also mentions Einstein's experiment using two simultaneous light flashes to prove the concept of simultaneity and how this relates to time dilation.
  • #36
Perhaps 4Newton has in mind a frame where the CBR is isotropic.

When the issue of the universal applicability of SR arises, it is almost always defended by resort to its postulated tenants. But these postulates are the root of the contention. By way of example, we can question whether all inertial frames are equal in the sense that every experiment will yield the same result - measurement of the dipole CBR anisotrophy is clearly not the same at different times of the year - (the Earth's orbital tangent velocity very nearly approximates an inertial frame ) so the most fundamental belief upon which Einstein founded his theory (that it is not possible to detect velocity wrt space), is questionable.
 
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  • #37
4Newton said:
I have no objections to relativity. All I am or have been doing is analyzing the nature of relativity and trying to pick out facts of the relationship of space-time in view of relativity.
This is surprising, because some of the things you've said seem to contradict relativity.

4Newton said:
Are not all lines in Minkowski space-time world lines, BD, AF, or ED.
Every curve can be called a world line, but only a straight line can be the world line of an inertial observer. The line also has to be "timelike". A line like BF (not drawn in the diagram) can't be the world line of an actual physical observer because that guy would have to move faster than the speed of light.

4Newton said:
Are you saying that all time measured no matter what you path, is proper time?
I suppose that you could put it that way, but what I'm really saying is that a clock measures the proper time along the clock's world line (almost no matter what that world line looks like; it has to be timelike, but it doesn't have to be straight).

4Newton said:
Now this is where we run into a problem. If you take a clock and go from B to E your clock will tick at a slower rate.
It will tick at a slower rate to you (assuming that your world line is AD), but it will always tick at the same rate to me (assuming that I'm moving with the clock).

4Newton said:
If the terms, real time, proper time, world line, clock time, rest frame, zero rest frame, can not be used to indicate the difference then it would be useful if the different paths and time over the different paths could be referred to by some words that indicated the differences. If you have any words that describe these differences I would be very happy if anyone would tell me the correct terms to use.
I don't know that "real time" would mean, unless it's a synonym for "proper time". I think I've made it clear enough what "proper time" and "world line" means. (Let me know if you disagree). "clock time"... Isn't that just another synonym for "proper time"? The spacetime diagram represents the "rest frame" of an inertial observer. His world line in this diagram is AD. I don't know what a "zero rest frame" is, unless it just means "rest frame".
 
  • #38
Where does it end, 4Newton? You've had this same discussion here before and it didn't take you where you wanted it to. I'd imagine you've had the same discussion in other places before you came here and had the same unsatisfactory result. Is there ever a point where you've repeated the same discussion enough times to re-evaluate your position in it?
 
  • #39
yogi said:
Perhaps 4Newton has in mind a frame where the CBR is isotropic.

When the issue of the universal applicability of SR arises, it is almost always defended by resort to its postulated tenants. But these postulates are the root of the contention. By way of example, we can question whether all inertial frames are equal in the sense that every experiment will yield the same result - measurement of the dipole CBR anisotrophy is clearly not the same at different times of the year - (the Earth's orbital tangent velocity very nearly approximates an inertial frame ) so the most fundamental belief upon which Einstein founded his theory (that it is not possible to detect velocity wrt space), is questionable.
You are wrong, yogi, in that the CMB does not provide a frame in which the laws of the universe are different from everywhere else.

You are correct that the Earth has a measurable velocity through the CMB, and due to our various orbiting motions, that velocity changes - but that fact does not affect our local laws of physics. For exampe an MM experiment performed on earth: If the CMB were really the classical aether, the MM experiment would show it.
 
  • #40
This is surprising, because some of the things you've said seem to contradict relativity.
Your term “seem” to contradict relativity is the major problem I am having on this topic. No one appears to be knowledgeable enough on the subject of SR to point out where there is any contradiction with relativity. Every one just states that it is wrong and then throws out some terms that are suppose to prove it. When I ask what they mean by the terms you see what happens.

Every curve can be called a world line, but only a straight line can be the world line of an inertial observer. The line also has to be "timelike". A line like BF (not drawn in the diagram) can't be the world line of an actual physical observer because that guy would have to move faster than the speed of light.
If every curve or straight line is a world line then that term has little meaning when trying to describe the difference between a clock that moves between BD and one that moves between BE. By that definition both clocks move on world lines.
I do agree that line BF can not be the world line of an actual observer.
I suppose that you could put it that way, but what I'm really saying is that a clock measures the proper time along the clock's world line (almost no matter what that world line looks like; it has to be timelike, but it doesn't have to be straight).
This again is all well and good and I have no problem agreeing with that but it still does not help with the difference between clocks.
It will tick at a slower rate to you (assuming that your world line is AD), but it will always tick at the same rate to me (assuming that I'm moving with the clock).
I agree with this also, but the question still remains. A clock that goes from B to D by way of C shows more accumulated time than a clock that goes from B to D by way of E.. The clock that goes from B to C to D will always have the most accumulated time and will have ticked at the fastest possible rate. This is the basis for the twin paradox.
I don't know that "real time" would mean, unless it's a synonym for "proper time". I think I've made it clear enough what "proper time" and "world line" means. (Let me know if you disagree). "clock time"... Isn't that just another synonym for "proper time"? The space-time diagram represents the "rest frame" of an inertial observer. His world line in this diagram is AD. I don't know what a "zero rest frame" is, unless it just means "rest frame".
Now I think I see the difference between us. You seem to think that the drawing is that of a moving, inertial, frame. This can not be so. In this drawing according to Minkowski space-time line AD is where delta is (ct) and dx=0, dy=0, and dz=0 also your transition of light line AF would not be at a 45 degree angle if this was an inertial frame. The angle would be dependent on your spatial transition as seen by the angle between AF and BE.

Where does it end, 4Newton? You've had this same discussion here before and it didn't take you where you wanted it to. I'd imagine you've had the same discussion in other places before you came here and had the same unsatisfactory result. Is there ever a point where you've repeated the same discussion enough times to re-evaluate your position in it?
I think you can see that I have re-evaluated my understanding from our first discussion. I have gone back and re-studied SR I have analyzed Minkowski space-time and I have had some very satisfactory results in other postings. Why should it end? I am learning a lot. This is not the extent of the ideas I would like to present. I do however know that I must learn to present them in an acceptable manner. From the acceptance on other postings I can see that I am starting to learn. I can also see that from this thread I still seem to have some lack of ability to communicate.
 
  • #41
4Newton said:
Your term “seem” to contradict relativity is the major problem I am having on this topic. No one appears to be knowledgeable enough on the subject of SR to point out where there is any contradiction with relativity. Every one just states that it is wrong and then throws out some terms that are suppose to prove it. When I ask what they mean by the terms you see what happens.
The reason I said "seem" is that half the time I don't even understand what you're trying to say. Let's look at a few examples:

"If the speed of light is constant, then all points equal distant from the observer are simultaneous."

Huh?!? What does this mean? Distance in space? Distance in spacetime? If it's distance in spacetime, are you using the Euclidean metric or the Minkowski metric? Actually it doesn't matter much, because if you mean anything but distance in space, the statement is very wrong (unless you're just trying to say that from a light ray's perspective every event in spacetime happens at the same time), and if you mean distance in space...all points at any distance are of course simultaneous.

"Twin 1 must exceed the speed of light in space-time to arrive back on earth. If he did not exceed the speed of light in space-time he would never be able to return to Earth at the location that he left."

Are you talking about his four-velocity? The four-velocity is defined in a way that makes sure that it's magnitude is always 1 (actually c, but I'm using units in which c=1).


4Newton said:
By that definition both clocks move on world lines.
Everything moves on world lines. The world line of an object is just the path it takes through spacetime.

4Newton said:
This again is all well and good and I have no problem agreeing with that but it still does not help with the difference between clocks.
...
I agree with this also, but the question still remains. A clock that goes from B to D by way of C shows more accumulated time than a clock that goes from B to D by way of E.. The clock that goes from B to C to D will always have the most accumulated time and will have ticked at the fastest possible rate. This is the basis for the twin paradox.
I still don't understand what the problem is, and what question still remains.

4Newton said:
Now I think I see the difference between us. You seem to think that the drawing is that of a moving, inertial, frame.
Spacetime diagrams are supposed to represent the coordinates of an inertial observer. No inertial observer is any more or any less "moving" than any other.

4Newton said:
This can not be so. In this drawing according to Minkowski space-time line AD is where delta is (ct) and dx=0, dy=0, and dz=0
In this inertial frame (which represents the coordinates of the observer whose world line is AD), the events A and D have the same spatial coordinates, but different time coordinates. But to the observer whose world line is AE, A and D would not have the same spatial coordinates. To him the x coordinate of event A is 0 and the x coordinate of B is negative. (What is "stationary" in your frame is "moving to the left" in his frame).

4Newton said:
...also your transition of light line AF would not be at a 45 degree angle if this was an inertial frame. The angle would be dependent on your spatial transition as seen by the angle between AF and BE.
This is just wrong. The angle of the world line of a light ray is always 45 degrees, in every inertial frame.
 
  • #42
"If the speed of light is constant, then all points equal distant from the observer are simultaneous."

Huh?!? What does this mean? Distance in space? Distance in spacetime? If it's distance in spacetime, are you using the Euclidean metric or the Minkowski metric? Actually it doesn't matter much, because if you mean anything but distance in space, the statement is very wrong (unless you're just trying to say that from a light ray's perspective every event in spacetime happens at the same time), and if you mean distance in space...all points at any distance are of course simultaneous.

From post #6
Einstein used simultaneity to prove relativity.
In his experiment he used two simultaneous light flashes equal distant from an observer at rest with respect to the sources of the light. If the speed of light is constant, then all points equal distant from the observer are simultaneous. The simultaneous condition of the source may be established by synchronizing the two sources at the observer then moving them away each in the same manner to points that are equal distance.

Post #6 and others are given to show proof that all actions in our universe are simultaneous. The proof is from Einstein’s relativity. The basic reason for our disagreement is that I believe and I am showing proof of absolutes in our universe. When you start from a closed mindset that everything is relative and there are no absolutes you close your mind to understanding anything I am saying.

Your statement that, “all points at any distance in (space) are of course simultaneous”, is in agreement with me and the proof I gave. I think you will find some others about to disagree with you.

"Twin 1 must exceed the speed of light in space-time to arrive back on earth. If he did not exceed the speed of light in space-time he would never be able to return to Earth at the location that he left."

Are you talking about his four-velocity? The four-velocity is defined in a way that makes sure that it's magnitude is always 1 (actually c, but I'm using units in which c=1).

Because of possible disagreements over definitions I would prefer not to use any terms at this point. All that I am pointing out is that the length of line BD is shorter than the length of lines BED and transition along the vertical axis is at the same rate as transition on the horizontal axis as shown by the line AF speed of light.. Transition BD is at a rate equal to the speed of light. In order to remain in sync with BD any transition in the spatial dimension must exceed the speed of light in that space-time vector BE as indicated by the length of the lines.

I still don't understand what the problem is, and what question still remains.
The question is. Why do clocks indicate different accumulated time when brought back together after moving over different paths in space-time and how can anyone say that clocks measure real time when they disagree. Again looking for absolutes. It is obvious and I think agreed by all that all clocks keep time at the same rate and tick the fastest on all vertical lines that have no spatial transitions. Except of course AF if you could take that path your clock would always read zero.

Spacetime diagrams are supposed to represent the coordinates of an inertial observer. No inertial observer is any more or any less "moving" than any other.
Who made that rule? Minkowski space-time is not an inertial frame.

If you state that there is no condition that can have zero spatial transition show me some proof or reason of logic. Btw by definition is not proof.

The rest of you statements revolve around this same point. Minkowski space-time is real, absolute, it is not a relative viewpoint. Even if you assume that it is a relative viewpoint there is nothing that prohibits a viewpoint that has absolute zero spatial transition.

If you are able to go one direction at the speed of light limit and the opposite direction up to the same limit then by all rules of logic, experience, math... there must be a center point where the spatial transition is zero.
 
  • #43
4Newton said:
From post #6
Einstein used simultaneity to prove relativity.
In his experiment he used two simultaneous light flashes equal distant from an observer at rest with respect to the sources of the light. If the speed of light is constant, then all points equal distant from the observer are simultaneous. The simultaneous condition of the source may be established by synchronizing the two sources at the observer then moving them away each in the same manner to points that are equal distance.
This still doesn't make much sense to me. What do you mean by "all points equal distant from the observer"? Suppose that you add a y-axis to your spacetime diagram. x and y represents two spatial dimensions, and t the time dimension. Suppose also that the light signals are emitted at (0,-a,0) and (0,a,0), where the first coordinate is the time coordinate. Does "all points equal distant from the observer" mean the circle t=0, x²+y²=a², or does it mean the cylinder x²+y²=a²? Maybe it means something completely different. You really have to start specifying these things.

If you're talking about the circle, you have already restricted your attention to simultaneous events. It's always a bad idea to assume what you're trying to prove.

4Newton said:
Post #6 and others are given to show proof that all actions in our universe are simultaneous.
I don't even know what you're trying to say. To me it looks like all you have "proved" is that all events on any horizontal line in the spacetime diagram are simultanous (to the observer whose world line is AD), and this is obvious anyway since all points on such a line have the same time coordinate. (This is what we mean by "simultaneous". Two events are simultaneous in a frame if they have the same time coordinate in that frame).

4Newton said:
The proof is from Einstein’s relativity. The basic reason for our disagreement is that I believe and I am showing proof of absolutes in our universe. When you start from a closed mindset that everything is relative and there are no absolutes you close your mind to understanding anything I am saying.
You haven't proved anything yet. I don't even know what you're trying to do. I've tried to understand what you're trying to say, but I can't really make sense of it. My best guess is that this is what you have done:

You start with one observer's point of view and restrict your attention to a set of events that are simultaneous to him. Then you "prove" that those events are simultaneous to him (which is something you had already assumed, without mentioning it explicitly). Then you extend this result, without motivation, to other observers, and claim that all observers agree about what events are simultaneous. This would mean that there's an "absolute" rest frame, and that's your conclusion.

Am I wrong about this?

4Newton said:
Your statement that, “all points at any distance in (space) are of course simultaneous”, is in agreement with me and the proof I gave. I think you will find some others about to disagree with you.
Perhaps I didn't express myself clearly enough. First of all you have to remember that in SR, "the universe", "space", or whatever you want to call it, is just a horizontal line in the spacetime diagram. Different horizontal lines represent space at different times. But you also have to remember that you can't just assume that another observer will think of the same set of lines, as "lines of simultaneity", i.e. as lines that represent space at different times. If you open any good book about special relativity (I recommend the chapters about SR in "A first course in general relativity" by Bernard Schutz) you can find a proof of the fact that the postulate about the speed of light implies that the lines in your spacetime diagram that are simultaneity lines ("space at different times") to another observer, are not horizontal (unless his world line is exactly vertical).

To the observer whose world line is AD, distance "in space" means distance along a horizontal line. To an observer moving with velocity v, it means distance along a line with slope v.

There's more that I would like to comment but I have to go to bed. Maybe tomorrow.
 
  • #44
Russ - as I said above - when issues re SR arise - relativists always pontificate SR as an indisputable principle. - but you get exactly the same results when you apply the transforms to a fixed ether frame as did Lorentz. Or if you use Selleri transforms. Moreover, these alternatives intrinsically abrogate the twin or triplet paradox because the aging is not reciprocal. Einstein happened to be biased on the issue of whether velocity wrt space has any meaning. We now know that velocity wrt to the CBR does have meaning - Note that I didn't say that Galilean type frames where mechanical experiments are agreed to yield equal results would be affected by motion relative to the CBR - but experiemts attempting to measure one way isotrophy in free space might be a different story. .. This is the crux of SR, it is the one thing that distinguishes SR from the other theories that predict the same results for the experiments that have been made - and until that test has been made, SR continues to rest upon a leap of faith.
 
  • #45
In Minkowski space-time you are only able to show space as a single dimension. It is not possible to have all three spatial dimensions with 45-degree light lines. The speed of light limit is the vector sum of the three spatial dimensions. This is also true because you can only move in one direction at a time, in the spatial dimension, so although you have an unlimited degree of freedom to move in any direction any selected direction is perpendicular to the time dimension.

Time dimension is (ct) and the spatial dimension is (s) where (s) is any x,y,z, but ds is the vector sum of x,y,z.

I don't even know what you're trying to say. To me it looks like all you have "proved" is that all events on any horizontal line in the spacetime diagram are simultanous (to the observer whose world line is AD), and this is obvious anyway since all points on such a line have the same time coordinate. (This is what we mean by "simultaneous". Two events are simultaneous in a frame if they have the same time coordinate in that frame).

You still insist on referring to the drawing as a moving frame. It is not. This is an absolute coordinate system. All moving frames would lie on BE like lines. All vertical lines are at spatial positions without motion.

The exercise of simultaneous actions is to show proof as you have stated to validate the absolute Minkowski space-time drawing. In the simultaneous exercise equal distance is just that. Equal distance in all directions to form a sphere from a center point.

You start with one observer's point of view and restrict your attention to a set of events that are simultaneous to him. Then you "prove" that those events are simultaneous to him (which is something you had already assumed, without mentioning it explicitly). Then you extend this result, without motivation, to other observers, and claim that all observers agree about what events are simultaneous. This would mean that there's an "absolute" rest frame, and that's your conclusion.

I start with an undisputed simultaneous action, two flashing lights at the same point in space, I then move them apart with the observer at the center still having verification the two lights are simultaneous. I then state the obvious, that the change of the observers position changes his view of the lights flashing simultaneously but obviously his change of position does not change the action of the flashing lights.

I have thus separated the reality of observation from the reality of action. This is something SR failed to do.

It is also obvious that the flashing lights may be moved in any direction of any equal distance, line, circle or spherical, with the same results. This may then be done a second time. You now have an infinite number of simultaneous spheres intersecting another set of infinite number of simultaneous spheres resulting in the conclusion that all action is simultaneous through out all points of the universe.

I then go to the Minkowski space-time drawing, again this is absolute space-time with inertial frames moving on BE or ED like lines. As you have stated all horizontal lines are simultaneous time lines. Likewise all lines that the horizontal lines pass through must also be simultaneous time lines, but in action and observation moving frames fall out of sync. SR gives us the relationship of this function.

SR theory does not address functions outside of moving frames but interpretation has tried to extend and limit the concept of the universe to moving frames. This has had a negative result of hampering understanding of forces and dimensions.
 
  • #46
I've got a general comment:

It is (just barely) possible to correctly work problems in special relativity with a notion of "absolute time".

What one does is to pick some particular frame, and think of it as "special". One can then view the Lorentz contractions as actual contractions - the very atoms of matter are distorted, the time dilation as actual slowing of clocks, the synchronization differences between different observers as an "ether wind" due to motion relative to the special frame. This actually has one well-respected proponent, Bell (I don't recall his last name, he's the Bell who derived the Bell inequality).

However, usually people who attempt to work relativity with this mindset get sucked into a false claim that whatever frame they are picking as mentally special is actually special in some physical manner. This is not the case, for the "ether" here is just a mental construct. If one does the math, it turns out to be not physically detectible. The speed of light is constant for all observers in this version of relativity (just as it is for all others), and the laws of physics are constant as well. There is no measurement that will single out which frame is special. So one winds up with an infinite number of different notions of "absolute time", rather than one. There is no practical difference between the notion of an infinite number of "absolute times" and the notion that time is not absolute.

Exactly where some of the posters here fit into this spectrum is unclear. When I see people claiming at one point to be agreeing with relativity, and at another point railing against "relativisits", I conclude that either they are either really really, really confused, or not arguing in an honest manner, or both.
 
  • #47
pervect said:
It is (just barely) possible to correctly work problems in special relativity with a notion of "absolute time".

What one does is to pick some particular frame, and think of it as "special". One can then view the Lorentz contractions as actual contractions - the very atoms of matter are distorted, the time dilation as actual slowing of clocks, the synchronization differences between different observers as an "ether wind" due to motion relative to the special frame. This actually has one well-respected proponent, Bell (I don't recall his last name, he's the Bell who derived the Bell inequality).
John S. Bell argued that quantum theory and GR could be integrated into QG if a Lorentzian view of relativity was taken. i.e. 3+1D rather than 4D space-time. This required some form of preferred frame, 'absolute' might be too absolute a term for it! (Speakable and unspeakable in quantum mechanics)
Isham and Butterfield reasoned the same, or similar, argument, calling for a ‘preferred foliation of space-time. (2001, Physics meets Philosophy at the Planck Scale, ed. by C. Callender and N. Huggett. Cambridge University Press.)

I would argue too that if Mach’s Principle is brought into play then such a preferred foliation of space-time may be identified as selected by the distribution of mass and momentum in the universe, it is the one in which the CMB is globally isotropic.

A further conundrum: The Twin Paradox in a closed universe. If cosmic expansion slows down and reverses it would become hypothetically possible to circumnavigate the universe. So one twin stays put and the other goes off at 9.999…%c and eventually meets up with his/her brother/sister again. Has he/she has aged say 50 (say) years and his/her bother/sister has aged 10 billion years (say)? Or is it the other way round? How do you tell?

Just a thought - Garth
 
  • #48
4Newton said:
In order to remain in sync with BD any transition in the spatial dimension must exceed the speed of light in that space-time vector BE as indicated by the length of the lines.
I think I see what you mean now. I thought you were wrong about the definition of four-velocity, but you were talking about ordinary velocity. What you are wrong about is what the world line would look like. (It starts out almost vertical, but a little to the right, and then changes to almost vertical, but a little to the left. It can be almost vertical through the whole trip).

4Newton said:
Minkowski space-time is not an inertial frame.
That is correct. It's not a frame at all. It's a manifold. (If you really want to understand relativity you should study differential geometry).

4Newton said:
Even if you assume that it is a relative viewpoint there is nothing that prohibits a viewpoint that has absolute zero spatial transition.
Yes, there is. The idea of "absolute zero spatial transition" contradicts the speed of light postulate, time dilation, and any other relativistic effect you've ever heard of. We can talk more about this when you've done the excercise I suggest that you do at the end of this post.

4Newton said:
In Minkowski space-time you are only able to show space as a single dimension.
This is of course wrong. We can all visualize three dimensions, so you can draw two spatial dimensions ("the x-y plane") and the t axis. (I don't understand how you can say that we can't).

4Newton said:
It is not possible to have all three spatial dimensions with 45-degree light lines.
I hope that what you really mean is that we can't visualize all four dimensions in our minds (because there are three spatial dimensions in Minkowski space, and the world line of any light ray through the origin makes a 45° angle with all the spatial axes).

4Newton said:
The speed of light limit is the vector sum of the three spatial dimensions.
Wrong. Have you never heard of the "light cone"? Imagine a spacetime diagram with two spatial dimensions (because you won't be able to visualize one with three). The set {(t,x,y)|-t²+x²+y²=0} is the union of all lightlike worldlines through the origin. This is a cone, not a line. Hence the name "light cone".

4Newton said:
This is also true because you can only move in one direction at a time, in the spatial dimension, so although you have an unlimited degree of freedom to move in any direction any selected direction is perpendicular to the time dimension.
I can't tell if this is just wrong or "not even wrong".

4Newton said:
You still insist on referring to the drawing as a moving frame. It is not. This is an absolute coordinate system. All moving frames would lie on BE like lines. All vertical lines are at spatial positions without motion.
In that case, your spacetime diagram doesn't represent Minkowski space and has nothing to do with relativity, or the universe as we know it.

Do you understand that you can draw another spacetime diagram that represents the coordinates used by the observer whose world line is BE, and that in that diagram the line AD has a negative slope. In that frame AD is moving!

I also think you should make up your mind. There are no "absolute" coordinate systems in SR. In a previous post you said that you have no objections to relativity. Then why are you contradicting it?

4Newton said:
The exercise of simultaneous actions is to show proof as you have stated to validate the absolute Minkowski space-time drawing. In the simultaneous exercise equal distance is just that. Equal distance in all directions to form a sphere from a center point.
There are a few things you should realize:

1. In your spacetime diagram (with one spatial dimension), the "sphere" is just the two points (0,-a) and (0,a).
2. The events on this "sphere" are simultaneous by definition (in this frame), so it makes no sense at all to try to prove that they are simultaneous.
3. You're doing all of this from one observer's point of view, and that means that if you're able to prove a statement such as "these events are simultaneous" you have only proved that it's true in that observer's frame.

4Newton said:
I start with an undisputed simultaneous action, two flashing lights at the same point in space, I then move them apart with the observer at the center still having verification the two lights are simultaneous. I then state the obvious, that the change of the observers position changes his view of the lights flashing simultaneously but obviously his change of position does not change the action of the flashing lights.
This is correct (but pointless).

4Newton said:
I have thus separated the reality of observation from the reality of action. This is something SR failed to do.
You haven't separated anything that wasn't separated already.

That second sentence is so wrong I don't even know where to begin.

4Newton said:
It is also obvious that the flashing lights may be moved in any direction of any equal distance, line, circle or spherical, with the same results. This may then be done a second time. You now have an infinite number of simultaneous spheres intersecting another set of infinite number of simultaneous spheres resulting in the conclusion that all action is simultaneous through out all points of the universe.
So? You're just saying that (0,-a) and (0,a) are simultaneous regardless of what a is. This is nothing new.

If you're going to prove something, you can't begin with the assumption that what you're trying to prove is true, and than use that in your proof. No logical fallacy can be worse than that.

4Newton said:
...again this is absolute space-time with inertial frames moving on BE or ED like lines.
It seems pointless to continue this discussion unless you first do this excercise:

The observer whose world line is AE (extended to infinity in both directions) shoots a laser beam in the positive x direction at time -t (according to his own clock). It is reflected off a mirror that is perpendicular to the laser beam. At time t (according to the same clock) the reflected laser beam has returned, and hits the laser. Your task is to draw the path of the laser beam in your spacetime diagram. At what point in the spacetime diagram is the beam reflected? Can you figure out what time the AE observer's clock is displaying when the reflection event happens?
 
  • #49
Garth said:
John S. Bell argued that quantum theory and GR could be integrated into QG if a Lorentzian view of relativity was taken. i.e. 3+1D rather than 4D space-time. This required some form of preferred frame, 'absolute' might be too absolute a term for it! (Speakable and unspeakable in quantum mechanics)
Isham and Butterfield reasoned the same, or similar, argument, calling for a ‘preferred foliation of space-time. (2001, Physics meets Philosophy at the Planck Scale, ed. by C. Callender and N. Huggett. Cambridge University Press.)

Unfortunately, I just don't see any evidence for a preferred foiliation of space-time.

Most of the serious attempts I see to justify this preferred foiiliation involve long-range scalar fields of one form or other.

This seems to be more or less a requirement - electromagnetism has been very well studied and it's just not a good candidate for a preferred frame. In fact, Maxwell's equations are what led directly to relativity, once they were taken seriously enough. I.e. after the Michelson Morley experiments was performed it was realized that the conflict between Maxwell's equations and Newton's mechanics should be resolved in favor of Maxwell's equations. Later evidence has shown this to be undoubtedly the correct choice.

The weak force has already been unified mathematically with electromagnetism. (I suppose one might be able to break the theoretical unification with expirimental evidence, but no such evidence has come up.) The strong force might be a very distant possibility for creating some preferred frame, but I've never, ever seen any serious proposal that points to the strong force as the source of a "preferred frame".

This winds up with the need for a new, previously undiscovered force to create the preferred frame. But there are some fairly strict limits on the existence of the direct formes of such fields, ones that interact directly with matter.

Theories in which the new undiscovered field interacts only indirectly with matter are trickier. These theories usually can be described by their PPN predictions about gravity. To date, relativity has always been correct - only time will tell as we continue to test it further whether or not its predictions continue to be correct.
 
  • #50
Garth - good post. Pervect. I don't see the difficulty in conceptualizing - or for that matter, actualizing, preferred frames. Every frame where light exhibits one-way isotrophy can be considered an "at rest frame" -INDEPENDENT OF EINSTEIN'S SECOND POSTULATE). All such frames would also be CBR isotrophic. The test for SR or any questions with regard thereto, would be the measurement of one-way light paths in a frame that moves relative thereo (again we pretend that we have never heard of the second commandment (or excuse me "postulate") and proceed to conduct tests in this second frame which is in uniform motion wrt the first. We don't do this when we run tests in the Earth laboratory - we measure dilation wrt to Earth clocks in GPS and with high speed decaying particles, and with clocks flown in airplanes, but we do not test to see if the situation is truly reciprocal, and we have not made accurate enough tests in free space to determine whether the one way velocity of light is truly isotropic.

Note again, that while MLET (Modified Lorentz Ether Theory) gets the same answers as SR and by its nature resolves the Twin Paradox, it creates other issues that cannot be easily answered. But there are other transforms that rely upon the isotrophy of round trip light paths, rather than Einsteins second postulate, to arrive at the same results - and these theories are not encumbered by the physical affects that must occur to explain MLET
 
  • #51
pervect said:
Unfortunately, I just don't see any evidence for a preferred foiliation of space-time.

Most of the serious attempts I see to justify this preferred foiiliation involve long-range scalar fields of one form or other.

This seems to be more or less a requirement .
Exactly - you are describing the theory of Self Creation Cosmology. (plenty of posts on that in these Forums) A Brans Dicke type scalar field endows particles with inertial mass and interacts with the matter field.
pervect said:
This winds up with the need for a new, previously undiscovered force to create the preferred frame. But there are some fairly strict limits on the existence of the direct formes of such fields, ones that interact directly with matter.
The Equivalence Principle is violated in SCC - but in Eotvos type tests only to about one part in 10^-17 about three orders of magnitude less than present experimental sensitivity. However the scalar field exerts a force on all particles, equally to within this variability (10^-17), but not on photons which "fall" ar a rate 3/2 the Newtonian acceleration of gravity, this is testable but it has not been done to date.
pervect said:
Theories in which the new undiscovered field interacts only indirectly with matter are trickier. These theories usually can be described by their PPN predictions about gravity. To date, relativity has always been correct - only time will tell as we continue to test it further whether or not its predictions continue to be correct.
In all previous tests SCC predicts the same outcomes as GR. However we await the results of Gravity Probe B - the first experiment able to distinguish between them.

Garth
 
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  • #52
Fredrik:
I am preparing a set of drawings and the total concept in one document; it will include your last request, which I interpret to be a drawing of the path of light between inertial frames in my drawing . I think having more detailed and labeled drawings will make it easier to understand. I will try to include answers to all your other comments at one time.
 
  • #53
Excellent. I should probably tell you that I won't be able to visit this site tomorrow, so if you post it tomorrow you'll have to wait at least a day for a reply.
 
  • #54
Fredrik and all:

Here is a drawing that will illustrate space-time from an absolute viewpoint. Because as shown before there is a disagreement of terms used. I will use terms that are defined by the drawing. No other meaning or connotation is intended.

The drawing is a Minkowski space-time illustration.

Line AG and line HI are the lines of spatial rest frames. Spatial rest frames have transitions only in the time dimension. All vertical lines in the drawing are spatial rest frames.

Line AF shows the slope of the speed of light.

The spatial distance separates AG and HI AH. A light pulse is sent from B in frame AG to point E in frame HI and reflected back to D. This is the normal path of light that is being reflected back to the starting point. It is seen that the path of light is of equal length in both directions.

With the requirement that the speed of light is constant AF, the path of light is drawn between frames JL and MO. Both frames JL and MO are moving with a constant velocity indicated by the slope of JL and MO, in this case equal to one-half the speed of light. The distance AH = JM.. The path of light is JKL. It is seen that the distance the light must travel is longer going from J to K then the path of light in the case of B to E.. However the distance from K to L is shorter than the light path from E to D.. The total time over the entire path is Q-D longer than B to D

The difference between BED and JKL is as stated in Special Relativity where ( v ) is the slope JL and ( c ) is the slope AF.

The two paths of light in the MM experiment are represented by paths JKL and BPQ with JKL being in the direction of travel and BPQ perpendicular to the direction of travel. Note the total time of travel is the same for both paths. Both one way times however changes with velocity.

I hope from this you can see that Minkowski space-time is absolute and that this is in total agreement with SR.
 

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  • #55
4Newton said:
The drawing is a Minkowski space-time illustration.
I'm afraid not 4Newton! In your diagram you have drawn a Euclidean space not a Minkowski space. The sheet of paper that you used to draw the diagram is Euclidean and there is only so much that you can do with it to illustrate Lorentz transformations, well very little actually. You have to be very careful.

Basically a moving inertail frame of reference is not on a world-line that is simply a rotation of the observer's axes; otherwise Lorentz transformations would be cyclical, go fast enough (rotation by 2pi) and you would stand still!

Instead in a true Minkowski diagram a moving particle's world line asymtotically approaches the pi/4 AF dotted line.
4Newton said:
I hope from this you can see that Minkowski space-time is absolute and that this is in total agreement with SR.
Nice try but no! - In a true Minkowski diagram moving inertial observers are equivalent, there are no preferred frames of reference.

Garth
 
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  • #56
Garth:
I would be very interested in seeing your
Minkowski diagram a moving particle's world line asymptotically approaches the pi/4 AF dotted line.
 
  • #57
I do not have the means to send you a diagram.

However: Take your diagram, move J to A so the two systems have the same origin.

Now as J is moving relative to A his space and time axes have to be inclined at an angle to A's: J’s space axis is inclined upwards w.r.t. A’s.

On a Euclidean space diagram the space and the time axes are simply rotated, however on a Minkowski space diagram, on your diagram, J's space axis turns anti-clockwise and his time axis turns clockwise. From A's point of view J’s axes are no longer orthogonal but turned inwards.

However J's point of view is that it his own axes that are orthogonal, and it is A's axes that are 'turned inwards'.

As the axes of a 'moving' observer, as seen by a 'stationary' observer, turn inwards they asymptotically approach the light null world-line at pi/4 as the relative velocity approaches c.

The geometry of the light path reflected off your co-moving mirrors is the same in both frames of reference; they are equivalent.

I hope this helps, try drawing it again.
Garth
 
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  • #58
Garth:
As you rotate your axis of the moving frame what is your relationship to the speed of light line AF? Are you saying that you think the speed of light line also rotates?
 
  • #59
You're still missing the point 4Newton. Most of what you said in #54 is correct, but it has very little to do with relativity. The only events in your diagram where relativity is relevant is K and N, where light is reflected off a moving mirror without losing speed (like a tennis ball would). There is absolutely nothing in your diagram that gives any indication about what events are simultaneous to a moving observer.

Look at the drawing I attached. (It's ugly because I did it in Paint). The line marked t' is the world line of an observer who is moving to the right. (The line marked t is the world line of an observer that you would describe as stationary). The thin lines that change direction when they reach the line marked x' are light rays that the moving observer emits to the right at times -t3, -t2 and -t1. Mirrors have been placed at locations in space chosen so that the reflected light will return to the moving observer at times t1, t2 and t3. The moving observer would of course disagree with the time coordinates and say that the rays were emitted at times -t3', -t2' and -t1' but what's important is that he would agree that the times from emission to event O (the origin) is the same as the times from event O to detection (at times t1', t2' and t3').

Think about what this means. If light is emitted at time -t1' and is reflected at some unknown time and returns at time t1, then the "unknown" time of reflection must be t'=0. The reflection event has the same time coordinate as event O, and is therefore simultaneous with it.

This means that the entire line marked x' is a set of events that are simultaneous to the moving observer. Actually, any line that is parallell to x' is a simultaneity line for the moving observer.

This is not so strange as it might seem. If you have already accepted that the speed of light is the same to everyone, you shouldn't be surprised to see that the slice of spacetime that the moving observer calls "space" (the x' axis) is the one that puts the world line of a light ray exactly half-way between the x' axis and the t' axis.

Do you still think horizontal lines are events that are simultaneous to any observer?
 

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  • #60
I have studied you drawing and I have great difficulty understanding the concept you are trying to show. Your speed of light line I assume is the 45-degree line. Which also corresponds to your other light lines and of course the lines that are perpendicular to your light lines are also speed of light lines in the opposite directions. I can also understand your line t’ if you intend it to be a world line of a moving frame. All this is the same as my drawings. I do not understand what you line x’ is intended to be. If it is a world line it is to the right of the speed of light line and therefore anything moving on that line would be moving faster than the speed of light. I am certain this is not what you intend to say. It would help if you could relate how you drawing has anything to do with observation or reality.

In my drawing lines JL and MO are the same as AG and HI only they are rotated according to relativity. A light source moving on line JL sends out a pulse along the speed of light path JK and is reflected by a mirror at K and back to the light source that would than be at L. This is the same as the light path BED.

If you use this method to make a clock and the distance between the source and the mirror and back to the source in BDE is equal to the distance that light travels in one second 300 km. Then the light in the BED would take 1 second to return to the source.

Like wise you will notice that the distance between J and M is also equal to 300 km but this frame is moving in space, in this case one-half the speed of light. The total path of the light JKL is longer then the path BDE even though the distance between the source and the mirror is the same in both cases. The light arrives back at the source in JKL at a later time compared to BDE. The time difference varies asymptotically as the speed of the moving frame approaches the speed of light line.

This is relativity.

The line MNO is to show that the same is true no matter which direction the light travels.

I think it would help if you and Garth would go back and read Einstein’s paper.

It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.”

Translated by W. Perrett and G. B. Jeffery.

I covered simultaneity in previous post please review them and let me know what you don’t understand.

---
 
  • #61
4Newton said:
I have studied you drawing and I have great difficulty understanding the concept you are trying to show.
OK. I'll try to explain it.

4Newton said:
Your speed of light line I assume is the 45-degree line. Which also corresponds to your other light lines and of course the lines that are perpendicular to your light lines are also speed of light lines in the opposite directions.
That's exactly right.

4Newton said:
I can also understand your line t’ if you intend it to be a world line of a moving frame. All this is the same as my drawings.
Also correct. I would say that t' is the world line of an observer moving at constant velocity. (I'm not saying it's wrong to call it the world line of a moving frame, only that I prefer to call it what I called it).

4Newton said:
I do not understand what you line x’ is intended to be. If it is a world line it is to the right of the speed of light line and therefore anything moving on that line would be moving faster than the speed of light. I am certain this is not what you intend to say. It would help if you could relate how you drawing has anything to do with observation or reality.
This is the most important part of what I'm trying to say, so you shouldn't give up until you understand it. The line marked x' is obviously not the world line of an observer. As I said in my previous post, that line is a set of events that are simultaneous to the observer whose world line is the line marked t'. (All events on the line marked x' are simultaneous with the event O).

I also explained why we can be sure that those events are simultaneous in the moving observer's frame. If there is any particular part of that explanation that you think is difficult to understand, you should ask about that specific detail.

It's really hard to tell what you understand and what you don't understand. You seem to understand that the points in a spacetime diagram represent events, i.e. a location in space and time. To find the x coordinate of an event, just draw a vertical line through the event and see where that line intersects the x axis. To find the t coordinate of an event, just draw a horizontal line through the event and see where that line intersects the t axis. However, you don't seem to understand (or even care) how to find the coordinates that another observer would use. This is not something that can be ignored.

Any event that has coordinates (t,x) in a spactime diagram will have have different coordinates (t',x') in the frame of an observer who's moving relative to the "spatial origin" (the point in space that's represented by the t axis in the diagram). It is easy to find the set of events that have x'=0. This is just the world line of the observer. This is the line I marked t' in my drawing. It is more difficult to find the set of events that have t'=0 (i.e. the x' axis), but the trick I used is sufficient to find them. Since we know that the speed of light is the same to all observers, we know that if the moving observer emits light to the right when his clock displays 4:59:48, and that the light is reflected by a mirror and returns to the observer when his clock displays 5:00:02, the reflection must have happened when the clock was at 5:00:00.

It is obvious from the diagram that the reflection must happen on the line marked x'. This means that this line is the x' axis, i.e. the line where t'=0. This line consists of all events that are simultaneous with the event at the origin of the diagram. This line is what the moving observer would call "space, at time t'=0".

4Newton said:
In my drawing lines JL and MO are the same as AG and HI only they are rotated according to relativity.
You don't seem to realize that this rotation is exactly the same in non-relativistic theory. This part of what you're saying has nothing to do with relativity.

It seems pointless to comment your diagrams any further until you've understood what I'm saying about simultaneity.

4Newton said:
I think it would help if you and Garth would go back and read Einstein’s paper.
You really have to stop assuming that you are right and everyone else is wrong. Garth, pervect and I understand special relativty. You don't. Not yet anyway, but you can learn if you listen.

4Newton said:
I covered simultaneity in previous post please review them and let me know what you don’t understand.
I already have. It's often hard to tell if what you're saying is wrong or "not even wrong" because you sometimes don't make sense. The biggest problem is that you're implicitly assuming that what you're trying to prove is true, which is the worst mistake you could possibly make.

What would you say if I made a diagram that consists of one vertical line, and one horizontal line, that represent two spatial dimensions, and said that this diagram proves that the concepts of left and right are absolute? Would you accept that as proof? Of course not. Your proof is flawed in exactly the same way as this one that I just made up.
 
  • #62
I think I see what your drawing is saying. You are talking about light synchronization of clocks. As shown in:
http://homepage.sunrise.ch/homepage/schatzer/space-time.html

The only problem you have is that you can have more than one light synchronized clock in the same moving frame. The many different possible clocks will not agree with each other.

If you look at my drawing GST-02 you will see that in an absolute spatial rest frame light will transition in all directions that are equal distance from the source in the same amount of time X0 to X1 or X1 back to X0 take the same amount of time. T1-T0 = T2-T1. If the light starts at X0 at 3:58 and arrives back at 4:02 then the clock is at the reflection point is 4:00.

In a moving frame X’0 the light does not transition the same path in all directions that are equal distance to the source. The time to the reflection point will be the fastest perpendicular to the direction of travel and the time to the reflection point will be the longest in the direction of travel as shown in GST-02.

The difference between a spatial rest frame X0 and a frame moving at one half the speed of light X’0 is shown in the drawing GST-02. A light path, X’0 to T’3, that is perpendicular to the direction of travel is 1.2 times longer then the path from X0 to T1. A light path that is in the direction of travel X’0 to X’1 is 1.7 times longer.

The question is; what if any is the real time.
You really have to stop assuming that you are right and everyone else is wrong. Garth, pervect and I understand special relativty. You don't. Not yet anyway, but you can learn if you listen.
I think you should consider that this cuts both ways. The laws of physics are not elected by popular vote.
I already have. It's often hard to tell if what you're saying is wrong or "not even wrong" because you sometimes don't make sense. The biggest problem is that you're implicitly assuming that what you're trying to prove is true, which is the worst mistake you could possibly make.
I don’t think you are trying to understand what I am saying. You have never questioned any particular point that you say does not make sense.

I am not trying to prove anything. I am simply trying to resolve contradictions and suppose paradoxes in theories of the universe.

You make the assumption that your understanding of the theories is absolutely correct in spite of the contradictions and paradoxes. Your solutions are that we just need to accept concepts that make no sense. How do you resolve the fact that clocks do not agree and yet you say that clocks tell real time? You assume that clock time is absolute and that all of nature must bend to match that concept.

I only look to see if there is a more global understanding of space-time that does not conflict with observation. I have posted a viewpoint based on accepted theory that resolves some of the conflicts. There is no circular logic here. The entire posting here is based on the accepted theory that the speed of light is an absolute constant and as a result the relationship of space and time is also absolute.
What would you say if I made a diagram that consists of one vertical line, and one horizontal line, that represent two spatial dimensions, and said that this diagram proves that the concepts of left and right are absolute? Would you accept that as proof? Of course not. Your proof is flawed in exactly the same way as this one that I just made up.
Not so, the difference is that even you accept that the speed of light is and absolute. And if you drew a line in your example above and said it is an absolute then you would be totally lacking if you could not show an absolute relation between the vertical and horizontal lines. You seem to be blind to seeing anything as absolute. Relativity does not mean that there are no absolute relationships.
 

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  • #63
4Newton said:
The only problem you have is that you can have more than one light synchronized clock in the same moving frame. The many different possible clocks will not agree with each other.
What do you mean by "in the same moving frame"? That doesn't make any sense. Do you mean that there are events that are not in the moving frame? In that case, what do you mean by "frame"?

4Newton said:
You make the assumption that your understanding of the theories is absolutely correct in spite of the contradictions and paradoxes.
There are no contradictions or paradoxes in SR.

Have you ever thought about what it would mean if there were? SR is just the mathematical theory of Minkowski space. If SR is logically inconsistent, then the definition of Minkowski space must be too. But Minkowski space is just [tex]\mathbb R^4[/tex] with a funny metric. The metric is just a function and couldn't possibly introduce any logical inconstencies into the theory. So if Minkowski space is inconsistent, the inconsistencies must come from the real numbers. But real numbers can be constructed from the rationals (as Dedekind cuts, or as equivalence classes of Cauchy sequences), and the rationals can be constructed from the integers. This means that if SR is inconsistent, the integers are too.

4Newton said:
Your solutions are that we just need to accept concepts that make no sense. How do you resolve the fact that clocks do not agree and yet you say that clocks tell real time? You assume that clock time is absolute and that all of nature must bend to match that concept.
But the concepts do make sense. The theory doesn't have to be wrong just because you don't understand it.

4Newton said:
There is no circular logic here. The entire posting here is based on the accepted theory that the speed of light is an absolute constant and as a result the relationship of space and time is also absolute.
What absolute relationship are you talking about? Would it be e.g. that the world lines of light are exactly half-way between the time axis and the space axis? Then why aren't they in your diagrams? The world line of a moving observer is his time axis. Yet you insist that space in a moving observer's frame is a horizontal line in your "absolute" diagram. If that's the case, then the world line of a light ray is closer to the time axis than to the space axis, so the speed of light in that frame can't be the same as in the "absolute" frame.

4Newton said:
If you look at my drawing GST-02...
Let's look at your drawing... The fifth diagram is the one that interests me. I'm sure you agree that x'=0 along the upper line. On what line would you say that t'=0?
 
  • #64
What do you mean by "in the same moving frame"? That doesn't make any sense. Do you mean that there are events that are not in the moving frame? In that case, what do you mean by frame"
http://plato.stanford.edu/entries/spacetime-iframes/
A “frame of reference” is a standard relative to which motion and rest may be measured; any set of points or objects that are at rest relative to one another enables us, in principle, to describe the relative motions of bodies. A frame of reference is therefore a purely kinematical device, for the geometrical description of motion without regard to the masses or forces involved. A dynamical account of motion leads to the idea of an “inertial frame,” or a reference frame relative to which motions have distinguished dynamical properties. For that reason an inertial frame has to be understood as a spatial reference frame together with some means of measuring time, so that uniform motions can be distinguished from accelerated motions.
The top right drawing fig. 1 (X0, X1, T0, T1, and T2) is a non-moving frame (frame with no spatial transition). I use the term moving and non-moving so it is not confused with non-inertial frames, (frames under acceleration).

The drawing fig. 4 (X’0, Z’1, T’3, and T’2) and fig. 5 (X’0, T’0, X’1, T’1, X’2 and T’2) are the lines of objects that are in the same inertial frame (moving in space at a transition equal to one half the speed of light). X’0 and T’2 are the same in both Fig. 4 and Fig. 5. Fig. 4 is the left view of Fig. 5 consistent with any engineering drawing.
There are no contradictions or paradoxes in SR.

But the concepts do make sense. The theory doesn't have to be wrong just because you don't understand it.
You have never heard of the Twin paradox? I have just given you an example of the speed of one way light clocks being different at the point of reflection in the same frame.

What absolute relationship are you talking about? Would it be e.g. that the world lines of light are exactly half-way between the time axis and the space axis? Then why aren't they in your diagrams? The world line of a moving observer is his time axis. Yet you insist that space in a moving observer's frame is a horizontal line in your "absolute" diagram. If that's the case, then the world line of a light ray is closer to the time axis than to the space axis, so the speed of light in that frame can't be the same as in the "absolute" frame.
The speed of light is constant. The relationship of the speed of light to spatial distance is always the same. The relationship of the speed of light to time is always the same. What you do, how you move, or what you think can never change that relationship. All actions of the smallest possible function occur at the same time throughout the whole universe as demonstrated in previous posts. As also demonstrated in previous posts no actions change the Now. All perceived changes are the result of viewpoint and not any change of the above relationship.

You seem unable to overcome you preset thoughts and look at things from a different viewpoint. You keep trying to look at everything from the viewpoint of an observer in a frame. The math will work no matter your viewpoint because all things are related but you need to have a relational formula for every event.

If you can open your view to a more global one it is then possible to examine the nature of other types of time keeping instruments and much more that you have never thought of.
Let's look at your drawing... The fifth diagram is the one that interests me. I'm sure you agree that x'=0 along the upper line. On what line would you say that t'=0?
I am not sure what you are asking. Maybe the above reply answers you question.
 
  • #65
The link to the page that talks about frames doesn't help. I know what a frame is. But I still don't know what you mean when you say things like "you can have more than one light synchronized clock in the same moving frame". That doesn't make sense. If a physical object like a clock is "in" one frame, than it's also in all the others.

A sentence like this one would make sense: "The object has a different shape in frame A than in frame B". But a sentence like this one makes no sense: "The object is in frame A".

4Newton said:
The drawing fig. 4 (X’0, Z’1, T’3, and T’2) and fig. 5 (X’0, T’0, X’1, T’1, X’2 and T’2) are the lines of objects that are in the same inertial frame (moving in space at a transition equal to one half the speed of light).
Here it seems that you believe that "in the same inertial frame" means "moving with the same velocity", but that would be wrong. When people (other than you) say things like "the velocity of this object in frame A is v" they don't mean that the object is any more in frame A than in frame B. They are only specifying in what frame the velocity is what they say it is, which is absolutely necessary, since the velocity (but not the presence of the object) depends on the choice of frame.

4Newton said:
You have never heard of the Twin paradox?
The "twin paradox" is not a paradox. When SR is used correctly the result is always that the astronaut twin is younger than his brother when he gets back. I have explained why in other posts in this forum. (You should try to understand what I'm trying to explain to you in this thread before you try to understand the twin "paradox").

4Newton said:
X’0 and T’2 are the same in both Fig. 4 and Fig. 5. Fig. 4 is the left view of Fig. 5
"The left view"?! I can only guess what you mean, because you're not making sense here either. I'm guessing that you mean that to an observer who's moving to the left in figure 4 (not drawn), the lines in figure 4 would look like in figure 5. That's not correct. The line X'0---Z'1 should be rotated counterclockwise by the same amount that the line X'0---T'2 has been rotated clockwise. This is absolutely necessary to ensure that the speed of light is the same in both frames (i.e. to both observers).

4Newton said:
The speed of light is constant.
Yes, it's certainly not accelerating. If you mean that it's the same to all different observers, I wonder why it isn't in your diagrams. (If it's c in figure 4, it's more than that in figure 5).

4Newton said:
The relationship of the speed of light to spatial distance is always the same. The relationship of the speed of light to time is always the same.
This is "not even wrong". (It's not clear what it means).

4Newton said:
As also demonstrated in previous posts no actions change the Now.
You haven't demonstrated anything.

4Newton said:
All perceived changes are the result of viewpoint and not any change of the above relationship.
I agree with that (if I understand you correctly), but all you've done is to choose an arbitrary observer's viewpoint and refuse to consider what the world looks like to any other observer. How can you talk about different viewpoints when you refuse to even consider the viewpoints of the observers you call "moving".

4Newton said:
You seem unable to overcome you preset thoughts and look at things from a different viewpoint.
That's not true. I'm just not going to abandon a logically consistent viewpoint, completely free from paradoxes and contradictions, for one that isn't.

4Newton said:
I am not sure what you are asking. Maybe the above reply answers you question.
No, it didn't. The reason I'm asking is that this is the most important concept in all of relativity (the one you need to understand to understand all the "paradoxes") and you don't seem to understand it.

I will ask the question again, in a different way. This time I have interpreted your figure 5 as representing the viewpoint of an observer who's moving to the left with speed v in figure 4, as the events in figure 4 occur.

Let's call the observer whose world line is the left vertical line in figure 4 "A" and the observer who's moving to the left in figure 4 (world line not drawn) "B". Suppose that observer A chooses to use the event at the lower left corner of figure 4 (where the vertical and the horizontal lines meet) as the origin of both space and time coordinates. Also suppose that B uses coordinates such that he is at x=0 all the time, and has t=0 when he passes the vertical line (in figure 4) that is the origin of spatial coordinates to A.

I would like to know which of these statements about figure 5 that you agree with, and which you disagree with.

1. The world line of observer B is a vertical line touching (T'0,X'0).
2. The world line of observer A is the line from (T'0,X'0) to (T'2,X'2).
3. Observer B's time axis (events with spatial coordinate 0 in frame B) is a vertical line touching (T'0,X'0).
4. Observer A's time axis (events with spatial coordinate 0 in frame A) is the line from (T'0,X'0) to (T'2,X'2), i.e. a line through (T'0,X'0) with slope 1/v.
5. Observer B's spatial axis (events with time coordinate 0 in frame B) is the horizontal line drawn in the figure.
6. Observer A's spatial axis (events with time coordinate 0 in frame A) is a line through (T'0,X'0) with slope v.
 
  • #66
The twin paradox...
Here is a real life and local example.
If you were to send your twin to Mercury for 1 Earth year, once the twin returns there would be a difference of roughly 1-3 minutes in difference in age. This may seem very small in electronic time but when you're talking larger distances it adds up rapidly and exponentially. This happens due to the speed of Mercury's orbit around the Sun and the gravitational interaction.

Age is relative...could you imagine living on Mercury(if it were possible), your life would be very very much shorter.
 
  • #67
Fredrik:
You seem to have a lot of problems understanding some very simple ideas. I did not know that you did not know how to read engineering drawings or what front left and right views were. I now present to you a 3D view maybe this will help. See drawing GST-06.

Fig. 1 is a simple stationary spatial ring. Around the ring is a reflecting surface. A light source at A sends a pulse of light out in all directions. The light pulse strikes the ring at time t1 and is reflected back to the source. All the reflections B, C, D… arrive back at the source E.

You will note that it takes the same the same amount of time for the light to travel from the source A , t0 to the ring, at t1 as it does for the light to travel from the ring at t1 back to the source at E, t2 and that the time of travel is the same in all directions.

You can also see that the reflection is midpoint from the start of the pulse at A to the return when the source is at E.

In Fig. 2 you see the same ring as above moving in a spatial direction to the right, right is the direction of tilt. You can now see that the light pulse sent out from the source when the source is at G is not reflected until J at time t4 in the direction of travel. However the light pulse in the direction perpendicular to the direction of travel is reflected at H at time t3. The time for the light to reach the reflection of the ring varies with the longest time in the direction of travel and the shortest opposite the direction of travel. The total round trip path of the light will always be the same in all directions.

The round trip time for the light pulse will vary dependent on the velocity of the ring and approach infinity as the slope of the line, representing velocity of the ring approaches the slope of the line that is the speed of light.

The round trip time for the light pulse will be minimum when the velocity of the ring is zero. As in Fig. 1.

All physical laws remain the same in a moving frame. Or what ever you want to call it, because transition causes corresponding physical changes that compensate. A mechanical clock will slow down because all mechanical clocks are dependent on mass of their elements to keep time. It is well established that mass increases with velocity. This increase in mass and the resulting slow down of the clock is in agreement with the round trip time of light in the same frame.

All other functions behave in the same manner.
 

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  • #68
Frigga:
Fredrik seems to agree that this takes place but he does not think it is a paradox. Everyone else realizes that if SR say that all physical laws are the same in all inertial frames and that your clock is based on physical laws. Then clocks after some time in one inertial frame disagree with clocks in another inertial frame contradict the first statement. To be nice we call that a paradox. SR is right in most cases just as Newton’s laws were right. It is these small “paradoxes” that should lead us to find a more true understanding of the universe. This is just as Einstein did with Newton.

All theories are open to expansion.
 
  • #69
In the absence of gravitational fields, i.e. curvature, SR is the appropriate theory. (So Mercury was not a good example).

The paradoxes of time, such as the Twin Paradox, are resolved by the understanding that simultaneity is also relative.

There is no paradox in the understanding that a clock in one inertial frame is observed to 'tick' at a different rate to a clock in another mutually moving inertial frame.

Things get interesting when curvature is introduced.

Garth
 
  • #70
You keep ignoring the things that prove you wrong. I will make one last attempt to explain this to you 4Newton, but if you keep ignoring what's important I will not waste any more time on you.

Look at http://w1.873.comhem.se/~u87325397/4Newton.jpg . It's just the drawing you did, with a few comments. Is there anything in it that you disagree with?

(In the drawing, I'm using primes to distiguish between the coordinates used by observer A and observer B. A is using the unprimed coordinates. B is using the primed coordinates).

Do you understand that different observers will have different opinions about what the time axis is, and that this implies that they also must have different opinions about what "slice" of spacetime they should call space?

If you put the line where t'=0 and y'=0 at any other place than where I have drawn it, A and B wouldn't agree about what the speed of light is.
 
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